Disclaimer: I am not a mathematician, I am a physicist.

The thermodynamic identity is usually expressed in the following differential form

$$ dU = TdS - PdV + \mu dN, $$

where $U$, $T$, $S$, $P$, $V$, $\mu$ and $N$ are the internal energy, temperature, entropy, pressure, volume, chemical potential and number of particles of the system respectively. If I am not mistaken, I can act with a vector, say $\frac{\partial}{\partial N}$, to yield

$$ \frac{\partial U}{\partial N} = T \frac{\partial S}{\partial N} - P \frac{\partial V}{\partial N} + \mu \implies \mu = \frac{\partial U}{\partial N} - T \frac{\partial S}{\partial N} + P \frac{\partial V}{\partial N}. $$

Consider the following question:

Consider a monoatomic ideal gas that lives at height $z$ above sea level, so each molecule has potential energy $mgz$ in addition to its kinetic energy. Show that the chemical potential $\mu$ is the same as if the gas were at sea level, plus am additional term $mgz$:

$$ \mu(z) = -k_b T \text{ln}\left[\frac{V}{N}\left(\frac{2\pi m k_bT}{h^2}\right)^{3/2}\right] + mgz. $$

My attempt was knowing that:

The "ideal monoatomic gas" implies $U = \frac{3}{2}k_bT$ (by equipartition theorem) and the validity of Sackur-Tetrode equation:

$$ S=k_bN\ln \left[{\frac {V}{N}}\left({\frac {4\pi m}{3h^{2}}}{\frac {U}{N}}\right)^{3/2}\right]+{\frac {5}{2}}, $$

together with the assumption that $V \neq V(N)$. If one uses the above formula for $\mu$ and takes the partial derivatives I yield

$$ \mu(z) = -k_b T \text{ln}\left[\frac{V}{N}\left(\frac{2\pi m k_bT}{h^2}\right)^{3/2}-\frac{3}{2}\right] + mgz, $$

which is almost correct except for that $-\frac{3}{2}$, although it still exhibits the problems described below.

I came to the conclusion that I don't know how to manipulate these equations in differential form, am I allowed to do the above "act with $\frac{\partial}{\partial N}$" business? The solution provided by the book is to say, hey hold $U$ and $V$ fixed so that the thermodynamic identity now reads

$$ 0 = TdS - 0 + \mu dN \implies \mu = T \left(\frac{\partial S}{\partial N}\right)_{V,U \text{ fixed}} $$

but $U = U(N)$, in particular $U = \frac{3}{2} k_b N T$ I could litterally make all the $N$s in $S$ dissapear by substituting $N = \frac{2 U}{3 k_b T}$ and claim that

$$ \mu = T \left(\frac{\partial S}{\partial N}\right)_{V,U \text{ fixed}} = 0, $$

which is ridiculous. I'm really lost with the mathematics behind this type of calculations... Which would be the correct way to proceed?


1 Answer 1


Start from the fact that $S=S(U,V,N)$ and the relation $dU = TdS - p dV + \mu dN$. As you say, holding $V$ and $U$ fixed, this yields

$$\mu = - T \left(\frac{\partial S}{\partial N}\right)_{V,U} = -T \lim_{h\rightarrow 0} \frac{S(U,V,N+h)-S(U,V,N)}{h}$$

Your claim is that if you write $U$ as a function of $N$, then you can make this zero by eliminating the explicit $N$ dependence of $S$. This isn't true, because you'd be introducing an implicit N-dependence via your expression for $U$.

To be more explicit, you are considering the following new function:

$$\Sigma(U,V,N) = S\left(U,V, \frac{2U}{3kT}\right)$$ and saying that the $N$-dependence has disappeared. But you've forgotten about $T=T(U,V,N)$. Taking the partial derivative of this beast with respect to $N$ while holding $U$ and $V$ fixed would give you

$$\left(\frac{\partial \Sigma}{\partial N}\right)_{U,V} = \lim_{h\rightarrow 0}\frac{1}{h}\left[S\left(U,V,\frac{2U}{3kT(U,V,N+h)}\right) - S\left(U,V,\frac{2U}{3kT(U,V,N)}\right)\right]$$

This is, of course, just the chain rule:

$$\left(\frac{\partial \Sigma}{\partial N}\right)_{U,V} = \frac{\partial S}{\partial N}\left(U,V, \frac{2U}{3kT}\right) \cdot \frac{-2U}{3kT^2} \cdot \left(\frac{\partial T}{\partial N}\right)_{U,V} $$ $$= \frac{\partial S}{\partial N}\left(U,V, \frac{2U}{3kT}\right)\cdot \frac{-N}{T}\cdot \left(\frac{\partial T}{\partial N}\right)_{U,V}$$

But since $ N = \frac{2U}{3kT}$, we have

$$\left(\frac{\partial N}{\partial T}\right)_U = -\frac{2U}{3kT^2} = \frac{-N}{T}$$ $$\left(\frac{\partial T}{\partial N}\right)_U = -\frac{T}{N}$$

which leaves us with

$$\left(\frac{\partial \Sigma}{\partial N}\right)_{U,V} = \frac{\partial S}{\partial N} \left(U,V, \frac{2U}{3kT}\right)$$

as we would expect.

  • $\begingroup$ Surely then I should view $\left(\frac{\partial \Sigma}{\partial N}\right)_{U, V}$ rather than as a partial derivative, as a total derivative holding $U$ and $V$ fixed. Surely I have a naive understanding of partial derivatives, but whenever I see one I treat everything that is not the differentiating variable as a constant. $\endgroup$ Commented Apr 21, 2020 at 22:20
  • $\begingroup$ Wait a minute. I think I get it, you only treat as constants the variables listed in the partial derivative, implying that the rest of the variables (such as $T$) are to be taken, if possible, as a function of the differentiating variable in question (in our case $N$). That would at least explain why you didnt apply the chain rule for $U$ but did for $T$, cause is not being held constant. Is this conclusion correct? $\endgroup$ Commented Apr 21, 2020 at 22:24
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    $\begingroup$ @FriendlyLagrangian In the present discussion, there are three independent variables $(U,V,N)$ and then several functions of them - the entropy $S$, and its derivatives $T,p,$ and $\mu$. When you wrote that $U = \frac{3}{2}NkT$, you imposed a constraint - effectively setting $T(U,V,N) = \frac{2U}{3Nk}$. $\endgroup$
    – J. Murray
    Commented Apr 21, 2020 at 22:31
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    $\begingroup$ @FriendlyLagrangian As long as that constraint is assumed to be satsified, you can freely substitute the variable $N$ with the expression $\frac{2U}{3kT(U,V,N)}$. However, the partial derivative of each (holding $U$ and $V$ fixed) is the same (obviously, or the expressions would not be equal to one another). $\endgroup$
    – J. Murray
    Commented Apr 21, 2020 at 22:35
  • 1
    $\begingroup$ @FriendlyLagrangian Yes, it sounds like you have the right idea. As I said, in my answer I treated $U,V,$ and $N$ as independent variables and everything else as functions of those quantities. Partial derivatives are then taken by holding two of those variables fixed and varying the other. If you want to switch to a different point of view - in which for instance the independent variables are $U,p,$ and $N$ - then you would perform a Legendre transformation. $\endgroup$
    – J. Murray
    Commented Apr 21, 2020 at 23:13

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